Solving the PDE $\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial ^2 \phi}{\partial y^2} = K \phi$ using the separation of variables

Question

I'm trying to solve $\dfrac{\partial^2 \phi}{\partial x^2} + \dfrac{\partial ^2 \phi}{\partial y^2} = K \phi$ with $K$ constant

I let $\phi = XY$ then got $X''/X + Y''/Y = K$ but I'm not sure where to go here

Answer 1

I'm a little bit rusty in this, so I hope I'm not doing any foolish mistakes.

Assume $\phi = X(x)Y(y)$, then as you already calculated: $$ Y(y)X''(x) + Y''(y)X(x) = KX(x)Y(y) $$ Assuming neither $X$ nor $Y$ vanish in their domain, this means that, $$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = K $$ From which we get that $$ \frac{X''(x)}{X(x)} = K-\frac{Y''(y)}{Y(y)} $$ Since each side depends on different variables, this equality holds independently of the value of $x$ or $y$. Then each side must be constant! (it doesn't depend on the value of $x$ or $y$!) So, $$ \frac{X''(x)}{X(x)} = \lambda = K-\frac{Y''(y)}{Y(y)} $$ Then, as it's already been done in the comments: $$ \frac{X''(x)}{X(x)} = \lambda \quad\text{and}\quad \frac{Y''(y)}{Y(y)} = \lambda - K = \lambda' $$ The solution to this ODE depends on $\lambda$, but essentially, if $\lambda\neq 0$ we get $$X(x) = Ae^{\sqrt{\lambda} x} + Be^{-\sqrt{\lambda} x}$$ and if $\lambda= 0$ we get $X(x) =Ax+B$.

It's also important to observe that the solution you get (aside from depending on the sign of $\lambda$) is only defined on the interval where neither $X$ or $Y$ vanishes! But you may be able to extend that solution to the whole domain.